Understanding E Multiplication: Solving Odd Equations with Exponents

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In summary, the equation \frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm 2e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }} is true and follows the rule of exponents for exponential functions. The missing factor of 2 in the numerator was corrected to show that the equation simplifies to 2(e^{2t})^2 = 2e^{4t}. Additionally, the mistake of (e^{2t})^{2t} instead of (e^{2t})^
  • #1
Pengwuino
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[tex]
\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm 2e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}
[/tex]
Is that true? What am i not realizing here...
 
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  • #2
Pengwuino said:
[tex]
\frac{{{\rm e}^{{\rm 2t}} }}{{\sqrt {e^{4t} + t} }}*2e^{2t} = \frac{{{\rm e}^{{\rm 4t}} }}{{\sqrt {e^{4t} + t} }}
[/tex]
Is that true? What am i not realizing here...

you left out the factor of 2 in the numerator
 
  • #3
oops yah i did... ill fix that

but is that true? I can't seem to grasp it... and its midnight so i don't think i will be able to easily realize what's going on

I was thinking that it should just be 3e^2t... and I don't understand how it just got turned into e^4t
 
  • #4
Pengwuino said:
oops yah i did... ill fix that
but is that true? I can't seem to grasp it... and its midnight so i don't think i will be able to easily realize what's going on
I was thinking that it should just be 3e^2t... and I don't understand how it just got turned into e^4t
No, [tex]e^{2t} + 2 e^{2t}[/tex] would be equal to [tex]3 e^{2t}[/tex].
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^2 = 2 e^{4t}[/tex]

Or for an even more straight forward example,
x + 2x = 3x
x * 2x = 2x^2
 
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  • #5
youre lucky though, since the [tex]e^x[/tex] function was proven to be exponential, thus it follows the rules of exponents, just like anything else. the rest is just algebra.
[tex]2^2 * 2^2=4 * 4=16=2^4[/tex]
or more generally
[tex]x^a * x^b=x^{ab}[/tex] for all bases that correspond to exponential functions. since the "e" function is exponential, you can use these laws.
have fun
 
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  • #6
there is a square missing:
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]
 
  • #7
gerben said:
there is a square missing:
[tex]e^{2t} * 2 e^{2t} = 2 (e^{2t})^{2t} = 2 e^{4t^2}[/tex]

[tex]e^{2t}* e^{2t}=e^{2t+2t}[/tex], not [tex](e^{2t})^{2t}[/tex]
 
  • #8
Pengwuino said:
oops yah i did... ill fix that

but is that true? I can't seem to grasp it... and its midnight so i don't think i will be able to easily realize what's going on

I was thinking that it should just be 3e^2t... and I don't understand how it just got turned into e^4t

lol i thought that's what the problem was, where did the 2 go? :tongue2:
 

1. Is there a specific method for multiplying out oddly?

Yes, there are multiple methods for multiplying out oddly depending on the specific expression. It is important to follow the order of operations and use appropriate algebraic rules.

2. How do I know if I have multiplied out an expression correctly?

You can check your work by simplifying the expression and comparing it to the original. Make sure to distribute any coefficients and combine like terms.

3. Can you provide an example of multiplying out oddly?

Sure, an example would be (2x + 3)(x + 5). After distributing the 2x and 3, and then combining like terms, the expression becomes 2x^2 + 13x + 15.

4. Are there any common mistakes to avoid when multiplying out oddly?

Yes, a common mistake is forgetting to distribute coefficients or not following the correct order of operations. It is also important to pay attention to signs and make sure to combine like terms correctly.

5. Is there a difference between multiplying out oddly and multiplying out evenly?

Yes, multiplying out oddly involves expressions with at least one odd exponent, while multiplying out evenly involves expressions with even exponents. The methods for multiplying out these expressions may differ slightly.

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